[" n-factor of "FeC_(2)O_(4)" in the change: "],[FeC_(2)O_(4)rarr Fe_(3)++CO_(2)" is : "]

Equivalent mass of FeC_(2)O_4 in the change, FeC_(2)O_(4) rarr Fe^(3+) + CO_(2) is

The equivalent weight of FeC_(2)O_(4) in the change FeC_(2)O_(4)rarrFe^(3+)+CO_(2) is

The equivalent weight of FeC_(2)O_(4) in the change FeC_(2)O_(4)rarrFe^(3+)+CO_(2) is

Equivalent weight of FeC_2O_4 in the change : FeC_2O_4 rarr Fe^(3+) + CO_2 is :

Equivalent mass of a substance may be calculated as, Equivalent mass =("Molecular mass")/("n-factor")=("Atomic mass")/(n-factor") n-factor= Basicity of acid or acidity of base n-factor= Number of moles of electrons gainted or lost per mole of oxidising or reducing agents n-factor= Total positive or negative valency of a salt n-factor= Valency of an ion. Concept of n-factor is very important for redox as well as for non-redox reactions. Equivalent mass of ferrous oxalate FeC_(2)O_(4) in the following reaction is : FeC_(2)O_(4) to Fe^(3+) + 2CO_(2)

Calculate the n-factor of reactants in the given reactions. (b) FeC_(2)O_(4)rarr Fe^(3+)+CO_(2)

Calculate the n-factor of reactants in the given reactions. (b) FeC_(2)O_(4)rarr Fe^(3+)+CO_(2)

Assertion: Equivalent weight of FeC_(2)O_(4) in the reaction, FeC_(2)O_(4)+ Oxidising agent rarr Fe^(3+)+CO_(2) is M//3 , where M is molar mass of FeC_(2)O_(4) . Reason: In theabove reaction, total two mole of electrons are given up by 1 mol e of FeC_(2)O_(4) to the oxidising agent.

Calculate the n-factor of reactants in the given chemical changes? (d) FeSO_(4)rarr Fe_(2)O_(3)